Certification Problem
Input (TPDB TRS_Standard/SK90/2.11)
The rewrite relation of the following TRS is considered.
|
+(0,y) |
→ |
y |
(1) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(2) |
|
-(0,y) |
→ |
0 |
(3) |
|
-(x,0) |
→ |
x |
(4) |
|
-(s(x),s(y)) |
→ |
-(x,y) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(-) |
= |
2 |
|
weight(-) |
= |
0 |
|
|
|
| prec(s) |
= |
0 |
|
weight(s) |
= |
1 |
|
|
|
| prec(+) |
= |
1 |
|
weight(+) |
= |
0 |
|
|
|
| prec(0) |
= |
3 |
|
weight(0) |
= |
4 |
|
|
|
all of the following rules can be deleted.
|
+(0,y) |
→ |
y |
(1) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(2) |
|
-(0,y) |
→ |
0 |
(3) |
|
-(x,0) |
→ |
x |
(4) |
|
-(s(x),s(y)) |
→ |
-(x,y) |
(5) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.